To my understanding, most of the qubits we use today are just Josephson junctions coupled to a resonator that is triggering the qubits to go to different states by using microwave pulses. The same resonators are also used to read the information form qubits. As shown in this picture:Quantum COmputer chip So I was wondering how does a qubit readout, when we measure it, gets read and how it is interpreted. How does raw information from qubit look like?

  • $\begingroup$ There are at least as many readout strategies as there are qubit types. Are you asking about readout in general or is the question specific to the transmon in your example? $\endgroup$ Commented May 19, 2019 at 3:19
  • 2
    $\begingroup$ The author used the word "most", so I gave the most common type of readout for superconducting qubits used today. $\endgroup$
    – psitae
    Commented May 19, 2019 at 6:06
  • $\begingroup$ @OferNaaman I am thinking of superconducting qubits most of the companies use and develop. Other technologies like trapped ions are less common. $\endgroup$ Commented May 19, 2019 at 6:53

1 Answer 1


What follows turned out to be a rather technical explanation, so I'll start with the main point: The qubit state can change the resonator's state, and the resonator's state can be easily measured only if there is a large different in frequencies between the qubit and the resonator.

Let's model a qubit as a two-level system and a resonator as a harmonic oscillator. We need to give both a characteristic frequency, so let's call the qubit frequency $\omega_q$ and the resonator frequency $\omega_r$. We also need to characterize the strength of the interaction (how fast energy is transferred from one thing to another), and that's usually called $g$.

Now we need to describe the dynamics of the qubit and resonator together. That's done by breaking the Hamiltonian down into three parts: qubit energy, resonator energy, and interaction strength.

$$H_{\text{qubit}} = \frac{1}{2} \hbar \omega_q \sigma_z $$

$$H_{\text{resonator}} = \hbar \omega_r \ a^\dagger a $$

$$H_I = \hbar g ( \sigma_+a + a^\dagger \sigma_-)$$

$$ H = H_{\text{qubit}} + H_{\text{resonator}} + H_I $$

This is the Jaynes-Cummings Hamiltonian. Note that the rotating wave approximation was used. Pauli operators are given by $\sigma$, and creation and annihilation operators are $a^\dagger$ and $a$.

Now, if the frequencies of the resonator and the qubit are the same (that's called the "resonance condition"), then they will exchange energy back and forth, much like an LC circuit or potential and kinetic energy in a pendulum. That's not ideal for measuring a qubit (you want to avoid as much as possible the measurement apparatus influencing the qubit state), so we make the frequency of the resonator very far away from the qubit's. This is a called a dispersive interaction.

Formally, if $\omega_r - \omega_q = \Delta$, the Hamiltonian (now in the interaction picture) can be expressed as

$$ H = \hbar \frac{g^2}{\Delta} ( |1\rangle \langle 1| + a^\dagger a \sigma_z ) $$

If the interaction between the qubit and the resonator is dispersive, and if the resonator is driven to a coherent state $|\alpha\rangle$, then the qubit states of $|0\rangle$ and $|1\rangle$ can be distinguished.

Examine the evolution of two initial states, $|0\rangle|\alpha\rangle$ and $|1\rangle|\alpha\rangle$:

$$ e^{-iHt/\hbar}|0\rangle|\alpha\rangle = |0\rangle|\alpha e^{-i\frac{g^2}{\Delta}t } \rangle $$

$$ e^{-iHt/\hbar}|1\rangle|\alpha\rangle = |1\rangle|\alpha e^{i\frac{g^2}{\Delta}t } \rangle $$

This is the point of the dispersive interaction: the coherent state of the resonator changes according to the state of the qubit, without having energy exchanged directly. You can see that the coherent state gets shifted counterclockwise for a 0 and clockwise for a 1. Furthermore, it's a simple matter to measure the state of a coherent resonator, since it can be described classically.

This is only an outline of the complete answer. I've assumed knowledge of basic quantum mechanics, changing reference frames (interaction picture, Schrodinger picture, etc), coherent states, and the rotating wave approximation. I also glossed over the derivation of the dispersive Hamiltonian, which you can find in "Introductory Quantum Optics" by Gerry and Knight.

  • $\begingroup$ Thank you very much for your answer! It is really helpful :) $\endgroup$ Commented May 19, 2019 at 6:54

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